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blondinia [14]
2 years ago
6

When using MakeCode Arcade, what is the easiest way to make modules?

Computers and Technology
1 answer:
IrinaK [193]2 years ago
5 0

When using MakeCode Arcade, the easiest way to make modules is: B. By creating functions.

<h3>What is a MakeCode Arcade?</h3>

A MakeCode Arcade can be defined as a web-based code editor that is designed and developed by Microsoft Inc., in order to avail its end users an ability to create retro-arcade games for dedicated computer hardware and the web.

Generally, software developers that code on MakeCode Arcade can either use JavaScript or modules (blocks) to build their software program (game)  in a web browser while using a grand total of eight (8) buttons such as:

  • Four (4) direction buttons.
  • A button
  • B button
  • Menu button
  • Reset button

<h3>What are modules?</h3>

Modules are also referred to as blocks and they can be defined as buttons, sprite or shakes that snap into each other to define the set of executable codes (program) that an Arcade will run.

In MakeCode Arcade, the easiest way to make modules is by creating functions because they reduce redundancy within the games.

Read more on functions here: brainly.com/question/15352352

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Answer:

#include<iostream>

using namespace std;

int main(){

   //initialization

  int arr1[10] = {2,4,6,1,7,9,0,3,5,8};

  int temp;

   int size_arr;

   //nested for loop

  for(int i=0;i<size_arr-1;i++){

   for(int j=0;j<size_arr-i-1;j++){

       if(arr1[j]>arr1[j+1]){ //compare

               //swapping

           temp = arr1[j];

           arr1[j]=arr1[j+1];

           arr1[j+1]=temp;

       }

   }

  }

  //display the each element

  for(int i=0;i<10;i++){

   cout<<arr1[i]<<" ";

  }

    return 0;

}

Explanation:

Create the main function and declare the variable and defining the array with 10 values.

take the nested for loop, nested loop means loop inside another loop.

the outer loop traverse in the array from 0 to size-1.

and inside loop traverse from 0 to size -i-1, because for every cycle of the outer loop the one element is sorted at the end. so, we do not consider the element from the last for every cycle of the outer loop. That's why (size-i-1)

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we take a third variable for swapping. after that, if the outer loop condition false it means all elements will traverse and then the loop will terminate.

and finally, take another for loop and display the output.

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